Solve for $x$, ignoring any extraneous solutions: $\dfrac{x^2 - 16}{x + 10} = \dfrac{-18x - 96}{x + 10}$
Multiply both sides by $x + 10$ $ \dfrac{x^2 - 16}{x + 10} (x + 10) = \dfrac{-18x - 96}{x + 10} (x + 10)$ $ x^2 - 16 = -18x - 96$ Subtract $-18x - 96$ from both sides: $ x^2 - 16 - (-18x - 96) = -18x - 96 - (-18x - 96)$ $ x^2 - 16 + 18x + 96 = 0$ $ x^2 + 80 + 18x = 0$ Factor the expression: $ (x + 8)(x + 10) = 0$ Therefore $x = -8$ or $x = -10$ However, the original expression is undefined when $x = -10$. Therefore, the only solution is $x = -8$.